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Question

If α and β are the roots of the equation x242kx+2e4lnk1=0 for some k and α2+β2=66, then α2+β2+α+β is equal to

A
66+82
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B
74
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C
6682
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D
58
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Solution

The correct option is A 66+82
We have,
α+β=42K
αβ=2k41
α2+β2=(α+β)22αβ
66=16×2k24k4+2.
k48k2+16=0
(k24)2=0
k2=4
k=±2, since k>0,k=+2
So, equation becomes x282x+31=0
α2+β2+α+β
=66+4k2
=66+82

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